Prove that, for all positive integral values of \(m\), \(x-1\) is a factor of \(x^m - 1\).

By the factor theorem, a polynomial \(p(x)\) has a linear factor \((x-a)\) if and only if \(p(a) = 0\).

If we take \(m\) to be a positive integer and set \(p(x) = x^m - 1\), then \(p(1) = 1^m - 1 = 0\) and therefore \((x-1)\) is a factor of \(x^m - 1\).

Moreover, we can write \[x^m - 1 = (x-1)(x^{m-1} + \dotsb + 1).\]

Prove that, if \(n\) is an integer greater than \(1\), then \(26^n - 26\) is an odd multiple of \(50\).

Since \(26^{n-2} + \dotsb + 26^1\) is an even integer, the bracketed factor is odd, and so \(13\) times the bracketed factor is also odd. That is, \(26^n - 26\) is an odd multiple of \(50\).

Hence, or otherwise, show that, if \(26^n\) is evaluated for any integral value of \(n\) greater than \(1\), then the last two digits of the result are independent of \(n\).

We can write \[26^n = \left( 26^n - 26 \right) + 26.\] By the second part of the question (which applies as \(n > 1\) is an integer), as the bracketed term is an odd multiple of \(50\), its final two digits will always be \(50\).

Thus, the final two digits of the right-hand side of the above will be \(76\); this is independent of the value of \(n\).